Question

The degree of dissociation is 0.4 at 400 K and 1.0 atm for the gaseous reaction $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$. Assuming ideal behaviour of all gases, calculate the density of equilibrium mixture at 400 K and 1.0 atmosphere. (Atomic mass of P = 31.0 and Cl = 35.5 )

• A. $4.12\times {10}^{-3}g/ml$
• B. $4.54\times {10}^{-3}g/ml$
• C. $3.52\times {10}^{-3}g/ml$
• D. None of these

Answer 1

Answer: (B)

$4.54\times {10}^{-3}g/ml$

Solution:- (B) $4.54\times {10}^{-3}gm/mL$

$\underset{\text{At equillibrium}}{\underset{\text{Initially}}{}}\underset{1-\alpha }{\underset{1}{P{Cl}_5}}\rightleftharpoons \underset{\alpha }{\underset{0}{P{Cl}_3}}+\underset{\alpha }{\underset{0}{{Cl}_2}}$

Therefore,

Total no. of moles at equillibrium $=1-\alpha +\alpha +\alpha =1+\alpha$

Given that $\alpha =0.4$

$\therefore$ Total no. of moles at equillibrium $=1+0.4=1.4$

As we know that,

$\frac{\text{Normal wt.}}{\text{Exp. wt.}}=1.4$

Normal wt. of $P{Cl}_5=208.5gm$

$\therefore \text{Exp. wt.}=\frac{208.5}{1.4}=148.93$

Now using ideal gas equation-

$PV=nRT$

$\because n=\frac{\text{Weight}\left(w\right)}{\text{Mol. wt.}\left(M\right)}$

$PV=\frac{w}{M}RT$

$\Rightarrow \frac{w}{V}=\frac{PM}{RT}$

As, $\text{Density}\left(d\right)=\frac{w}{V}$

$\therefore d=\frac{PM}{RT}$

$\Rightarrow d=\frac{1\times 148.93}{0.0821\times 400}=4.54gm/L=4.54\times {10}^{-3}gm/mL$

Hence the density of equilibrium mixture is $4.54\times {10}^{-3}gm/mL$.